2016 H2 Mathematics Paper 2 Question 7

Permutations and Combinations (P&C)

Answers

(i)

24{24}

(ii)

576{576}

(iii)

112{\frac{1}{12}}

(iv)

512{\frac{5}{12}}

Full solutions

(i)

Number of ways
=(43)×3!=24  ■\begin{align*} & = {4 \choose 3}\times 3! \\ & = 24 \; \blacksquare \end{align*}

(ii)

We will consider the complement, which consists of two cases
Complement case 1: no women/all men. Number of ways
=(63)×3!=120\begin{align*} & = {6 \choose 3}\times 3! \\ & = 120 \end{align*}
Complement case 2: no men/all women.
From (i), number of ways =24{=24}
Required number of ways
=(103)×3!−120−24=576  ■\begin{align*} & = {10 \choose 3}\times 3! - 120 - 24 \\ & = 576 \; \blacksquare \end{align*}

(iii)

Required probability
=7!×3!9!=112  ■\begin{align*} & = \frac{7! \times 3!}{9!} \\ & = \frac{1}{12} \; \blacksquare \end{align*}

(iv)

Required probability
=6!×(73)×3!9!=512  ■\begin{align*} & = \frac{6! \times {7 \choose 3} \times 3!}{9!} \\ & = \frac{5}{12} \; \blacksquare \end{align*}